Four numbers in increasing order form an $A.P.$ The sum of those numbers is $32$ and the ratio of the product of extremes to the product of means is $7: 15$. Find those numbers.

(2, 6, 10, 14) Let the four numbers in $A.P.$ be $(a-3d), (a-d), (a+d), (a+3d)$.
Given that the sum of the numbers is $32$:
$(a-3d) + (a-d) + (a+d) + (a+3d) = 32$
$4a = 32 \implies a = 8$.
The numbers are $(8-3d), (8-d), (8+d), (8+3d)$.
The product of extremes is $(8-3d)(8+3d) = 64 - 9d^2$.
The product of means is $(8-d)(8+d) = 64 - d^2$.
Given the ratio of the product of extremes to the product of means is $7:15$:
$\frac{64-9d^2}{64-d^2} = \frac{7}{15}$
$15(64-9d^2) = 7(64-d^2)$
$960 - 135d^2 = 448 - 7d^2$
$512 = 128d^2$
$d^2 = 4 \implies d = 2$ (since the numbers are in increasing order,$d > 0$).
Substituting $a=8$ and $d=2$:
$8-3(2) = 2$
$8-2 = 6$
$8+2 = 10$
$8+3(2) = 14$
The numbers are $2, 6, 10, 14$.